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Transformation Formulae — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae4 Marks
Question
Prove that: $\sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{11\text{x}}{2}=\sin2\text{x}\sin5\text{x}.$
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Answer

We have, $\text{LHS}=\sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{11\text{x}}{2}$ $=\ \frac{1}{2}\Big[2\sin\frac{7\text{x}}{2}\sin\frac{\text{x}}{2}+2\sin\frac{110}{2}\sin\frac{3\text{x}}{2}\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{7\text{x}}{2}-\frac{\text{x}}{2}\Big)-\cos\Big(\frac{7\text{x}}{2}+\frac{\text{x}}{2}\Big)+\cos\Big(\frac{110}{2}-\frac{3\text{x}}{2}\Big)-\cos\Big(\frac{110}{2}+\frac{3\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\frac{60}{2}-\cos\frac{80}{2}+\cos\frac{80}{2}-\cos\frac{140}{2}\Big]$ $=\ \frac{1}{2}[\cos3\text{x}-\cos4\text{x}+\cos4\text{x}-\cos7\text{x}]$ $=\ \frac{1}{2}[\cos3\text{x}-\cos7\text{x}]$ $=\ \frac{-1}{2}[\cos7\text{x}-\cos3\text{x}]$ $=\ \frac{-1}{2}\Big[-2\sin\Big(\frac{7\text{x}+3\text{x}}{2}\Big)\sin\Big(\frac{7\text{x}-3\text{x}}{2}\Big)\Big]$ $=\ \sin\frac{10\text{x}}{2}\sin\frac{4\text{x}}{2}$ $=\ \sin5\text{x}\sin2\text{x}$ $=\ \sin2\text{x}\sin5\text{x}$ $=\ \text{RHS}$ $\therefore\ \sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{110}{2}=\sin2\text{x}\sin5\text{x}.$ Hence proved.

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